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Course: Algebra (all content)>Unit 10

Lesson 25: Binomial theorem

Pascal's triangle and binomial expansion

Sal introduces Pascal's triangle, and shows how we can use it to figure out the coefficients in binomial expansions. Created by Sal Khan.

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• ¿What would happen when you have a negative term? Say (a-b)^3
• You can Make (a-b)^3 to (a+(-b))^3 and make the expantion
• Did you guys see this pattern? If i take the number 1.1 and raise it into those powers, i will get the same results of the Pascal's triangle. For example:
1.1^0 is equal to 1.
1.1^1 is equal to 1.1
1.1^2 is equal to 1.21
1.1^3 is equal to 1.331
1.1^4 is equal to 1.4641
And so on. Can anyone explain that?
• Yeah, I observed it when I first saw the Pascal’s triangle. It also works with 11. That’s because 11^n =(10+1)^n. And 1 raised to any power is always 1. So for 11^4 it is
(10^4) + (4*10^3*1^1)+(6*10^2*1^2)+(4*10*1^3)+10^0.

As you can see, the powers of 1 make no difference and the answer is simply 14641. Powers of 1.1 are just powers of 11/10 and so they follow the same pattern.
• Can't you just add the numbers like for example 2+1=3 so thats the next number under it and again and again cause while you were explaining all this i noticed this way
• Yes, it works that way too. Can you prove that they generate the same triangle?
• how was pascals triangle discovered?
• Where did this triangle come from?
• I've actually never heard Sal's explanation of the number of ways to get to each point in the triangle. An easier way to calculate the numbers is that each number is the sum of the two directly above it. The exceptions of course are all the 1s on the borders. The next row beyond what Sal wrote out is easy to calculate this way: it would be 1 5 10 10 5 1
• What does this Symbol mean ∑? THanks
• The symbol means "sum of."
• Is it just coincidence that the sum of the coefficients of (a+b)^n = 2^n? Does this have any connection to the open n-set of combinations?
• I will use aCb to mean a choose b. Notice that (a+b)^n expands as nC0 * a^n + nC1 * a^(n-1) * b + nC2 * a^(n-2) * b^2 + ... + nC0 * b^n. Now let a = b = 1. Notice that now all powers of a and b disappear and become ones, which don't affect the coefficients. But now, (a+b)^n is equal to (1+1)^n = 2^n. So we have 2^n = nC0 + nC1 + nC2 + ... + nC(n-1) + nCn. This proves that the sum of the coefficients is equal to 2^n.
• what if you have terms with exponents inside the brackets?
And also what do you do if there are multiple variables with exponents multiplied within the brackets?
I encountered this one reacently and became uncertain:
(3x^3y^2z)^4

How would that be solved and could you use pascals triangle for this?
Thank you:)
• First off, the example you provided wasn't a binomial. For your example, all you need are simple exponent rules.
If you meant to ask "what if there were multiple variables added/subtracted within the brackets" then you would use what is called Multinomial Theorem which is a generalized binomial theorem. When you are expanding a trinomial (3 variables) then you could actually use what is called "Pascal's pyramid" which is exactly what it sounds like.
You can read more on Pascal's pyramid and Multinomial Theorem here:
https://en.wikipedia.org/wiki/Multinomial_theorem
https://en.wikipedia.org/wiki/Pascal%27s_pyramid
• I have the equation (2x+1)^4, and it says to write in descending powers of x. What does this mean? I just did it normally using powers 1,4,6,4,1 and I got the right solution. Like this:
1((2x)^4)((1)^0) + 4((2x)^3)((1)^1) + 6((2x)^2)((1)^2) + 4((2x)^1)((1)^3) + 1((2x)1^0)((1)^4) = 16x^4 + 32x^2 + 8x +1
But the steps aren't the same. Instead, they took powers of 4,3,2,3,4 like this:
1((2x)^4)((1)^0) + 4((2x)^3))((1)^1) + 6((2x)^2)((1)^2) + 4((2x)^1)((1)^3) + (1)^4. They also get 16x^4 + 32x^3 + 24x^2 + 8x + 1.
Can someone please explain? Thank you!
• You have two errors in your calculations:

First: 4((2x)^3)((1)^1)
(2x)^3 = 8x^3
4*8x^3 = 32x^3
I think you made it into 32x^2

Second: You lost your term created by 6((2x)^2)((1)^2) = 24x^2

Hope this helps.